Consider a semiconductor junction between two different semiconductors. In the semiconductors page of Feynmann's lectures https://www.feynmanlectures.caltech.edu/III_14.html, it is stated that

"...the product $N_{p}N_{n}$ must be the same for the p-side as for the n-side... We have seen earlier, however, that this product depends only on the temperature and the gap energy of the crystal. Provided both sides of the crystal are at the same temperature, the two equations are consistent with the same value of the potential difference."

Here $N_{p}$ refers to the charge density of the positive charge carriers (e.g. the holes), while $N_{n}$ refers to the charge density of the negative charge carriers (e.g. the electrons) and it is stated earlier that $$ N_{p}N_{n}\propto e^{-E_{gap}/\kappa T} $$ where $E_{gap}$ is the band gap energy of the crystal, $\kappa$ is the Boltzmann constant and $T$ is temperature.

My question is what happens if the two semiconductors involved have a different band gap energy? How can this be consistent with the product of the of positive and negative charge carrier densities being the same either side of the junction?

Thanks in advance for any help.

  • 1
    $\begingroup$ Feynman is not discussion a heterojunction - the same semiconductor (doped differently) is on either side of the junction. In the case of a real heterojunction, there are a wide variety of possibilities of how the junctions line up, but $np$ will be different for the different materials. $\endgroup$
    – Jon Custer
    Aug 25, 2020 at 14:35
  • $\begingroup$ Notice the discussion you're quoting from starts with "We would like to discuss now what happens if we take two pieces of germanium or silicon ... and put them together to make a 'junction.'" --- as Jon says, we're talking about a junction of two pieces (or regions) of silicon or two pieces of germanium, not a junction between different semiconductors. $\endgroup$
    – The Photon
    Aug 25, 2020 at 14:55
  • $\begingroup$ It's not obvious to me why doping wouldn't affect the band gap energy. Am I to assume the doping is too minor for any change in the band gap energy? $\endgroup$
    – Chris
    Aug 25, 2020 at 15:05
  • $\begingroup$ @Chris - yes, typical doping levels are roughly one in a million. Very heavy doping leads to degenerate levels but few dopants can be made soluble at the required concentration without highly non-equilibrium processes. $\endgroup$
    – Jon Custer
    Aug 27, 2020 at 0:43

1 Answer 1


As others have pointed out in the comments, Feynman is not referring to junctions made of different materials, but rather, junctions having the same semiconductor with different doping on each side. Regarding the OP's last comment, if the material is not heavily doped, the band structure is not excessively modified, i.e. the band gap doesn't change, and the mass-action law $np=n_i^2$is still valid (the product $np$ is independent of the doping). The only effect the doping has on the band structure is to introduce allowed discrete energy states inside the band gap.


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